Let $A$ denote the quotient ring $\frac{\mathbb{Q}[X]}{\langle X^3 \rangle}$ then which of the following are true Let $A$ denote the quotient ring $\frac{\mathbb{Q}[X]}{\langle X^3 \rangle}$ then which of the following are true:
(A) There are exactly three distinct proper ideals in $A$
(B) There is only one prime ideal in $A$.
(C) $A$ is an integral domain.
(D) Let $f,g \in \mathbb{Q}[X]$ such that $\bar{f}\bar{g}=0$ in $A$, ($\bar{f},\bar{g}$ denotes the image of $f,g$ in $A$) then $f(0)g(0)=0$.
So here is my answer
(A) is true because $\langle X \rangle$, $\langle X^2 \rangle$ and $\langle X^3 \rangle = 0$ are the only proper ieals.
(B) is true because $\langle X \rangle$ is the only prime ideal.
(C) is not true because $\langle X^3 \rangle$ is not prime in  $\mathbb{Q}[X]$ hence $\frac{\mathbb{Q}[X]}{\langle X^3 \rangle}$ is not a domain.
(D) is true because the factors of $X^3$ are $X , X^2$.
Am I right in my assertions. Please verify!
 A: The conclusions for the first three are fine, but hopefully these are not the answers you are submitting to a teacher. They really need more justification than you've given. Here are some recommendations:
For (A): appeal to the fact that $\Bbb Q[X]$ is a principal ideal domain, and that the ideals containing $X^3$ correspond to the divisors of $X^3$.
For (B): There are so few ideals, so it is no problem to give explicit examples why $(X^3)$ and $(X^2)$ aren't prime. Then, argue that $\Bbb Q[X]/(X)$ is an integral domain to prove $(X)$ is prime.
For (C): You can appeal directly to your example given in (B) as to why it is not prime, and then this is airtight.
(D) Needs a lot more attention from you. What you have written is not a coherent justification by any means. You could, for example, argue that the hypothesis means that $X^3$ divides $f(X)g(X)$ in $\Bbb Q[X]$, and since $X$ is prime in $\Bbb Q[X]$, $X$ divides one of $f$ or $g$, say $f$. This should tell you immediately what $f(0)$ is, and it does not matter at all what $g(0)$ is at that point.
